As you move away from a lightsource

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As you move away from a light source, the wavelength of the light remains constant, but the observer may perceive a red shift due to the Doppler effect. This phenomenon occurs because light sources moving away from the observer appear to have longer wavelengths, while those moving towards them exhibit shorter wavelengths, or blue shift. Although the speed of light is constant, the frequency perceived by the observer decreases as the source moves away. The discussion clarifies that the physical properties of light do not change, but the perception of those properties can vary based on relative motion. Understanding these concepts is crucial in fields like astronomy and physics.
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As you move away from a lightsource...

Homework Statement


(from title)
...the wavelength of the light _______? (increase, stays the same, decrease, decrease in amplitude?)

Homework Equations





The Attempt at a Solution

 
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my guess would be the wavelength would stay the same, but I am not certain at all, anyone know?
 
explain why you believe it would stay the same.
 
well because wavelength is the distance between two successive points on waves right? well as you move away, the length dosent change, just the amount or possibly the frequency?
 
This is just a theory. Technically neither the frequency or wavelength change. But for the observer, if you account for the Doppler effect, light sources/stars moving away from the Earth appear to have a 'red shift' while those moving towards Earth have a 'blue shift'. Red light has a longer wavelength than blue light. Frequency appears to decrease, but speed of light remains the same.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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